let \[y=\sec^{-1}(x)\]
so \[\sec(y)=x\]
therefore \[y'\sec(y)\tan(y)=1\]
=> \[y'=\frac{1}{\sec(y)\tan(y)}\]
but we need this in terms of x
sec(y)=x remember!
and we can find tan(y) by looking at what sec(y) means
sec(y)=hyp/adj=x/1
so the missing side is opposite to y so we can find it by doing sqrt{x^2-1}
so we have
\[y'=\frac{1}{x*\frac{\sqrt{x^2-1}}{1}}=\frac{1}{x \sqrt{x^2-1}}\]